Quiz 1 Solutions. [1 point] No hackers are willing to waltz. Solution. x R(x) S(x) [1 point] No artists are unwilling to waltz.

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1 Massachusetts Institute of Technology Handout J/18.062J: Mathematics for Computer Science March Professors David Karger and Nancy Lynch Quiz 1 Solutions Problem 1 [10 points] Quantifiers Consider the following predicates where the universe of discourse is the set of all people: P (x): x is a hacker Q(x): x is an MIT student R(x): x is an artist S(x): x is willing to waltz Express each of the following statements using quantifiers, logical connectives, and and the above predicates: (a) [1 point] No hackers are willing to waltz. Solution. x P (x) S(x) (b) [1 point] No artists are unwilling to waltz. Solution. x R(x) S(x) (c) [1 point] All MIT students are hackers. Solution. x Q(x) P (x) (d) [1 point] MIT students are not artists. Solution. x Q(x) R(x) (e) [4 points] Does (d) follow logically from (a), (b), and (c)? If so, give a proof. If not, give a counterexample. Solution. Assume (a), (b), and (c). Then, we want to prove that, for any x such that Q(x) (for any MIT student), it follows that R(x) (that student is not an artist). From (c), we know that P (x) (that the student is a hacker). If we rearrange (a) to x P (x) S(x) then clearly S(x) (the student is not willing to waltz). So, if we rearrange (b) to x S(x) R(x) it follows that R(x) (the student is not an artist).

2 Handout 26: Quiz 1 Solutions 2 Name: (f) [1 point] Translate into English: ( x)(r(x) S(x) = Q(x)). Solution. All who are artists or people willing to waltz are MIT students. (g) [1 point] Translate into English: ( x)(r(x) Q(x)) = ( x)(p (x) = S(x)). Solution. If there is an artist who is not an MIT student, then all hackers are willing to waltz. Problem 2 [10 points] Induction Prove by induction (other methods will not receive credit) that (1 2) + (2 ) + ( 4) + + (n 1)n = 1 (n 1) n (n + 1) whenever n is a natural number greater than 1. Solution. Prove P (n) ::= (1 2) + (2 ) + ( 4) + + (n 1)n = 1 (n 1) n (n + 1) Basis. (P (2)) (1 2) = 1 (1) 2 (). Induction. Assume P (n). Prove P (n + 1). (1 2) + (2 ) + ( 4) + + (n 1)n }{{} Apply IH +n(n + 1) = 1 (n 1) n (n + 1) + n(n + 1) [ 1 = n(n + 1) = n(n + 1) ] (n 1) + 1 ] [ 1 n + 2 = 1 n(n + 1) [n + 2] Problem [10 points] Structural Induction Consider the sets S 1, S 2,... defined inductively as S 0 = {0, 1} For all n 1, S n = { x+y 2 x, y S n 1 } i.e., where S n contains the average values of each pair of numbers in S n 1. Note we allow x = y in the definition of S n. (a) [1 point] Write out the elements of each of the following sets: S 1, S 2, and S. Solution. S 1 {0, 1/2, 1} S 2 {0, 1/4, 1/2, /4, 1} S {0, 1/8, 1/4, /8, 1/2, 5/8, /4, 7/8, 1}

3 Handout 26: Quiz 1 Solutions Name: (b) [2 points] Generalize (in your own words or in symbols) to give the elements of S n. Solution. S n contains 2 n + 1 elements (including 0 and 1) evenly spaced along the unit interval; i.e., S n = {i/2 n 0 i 2 n }. (c) [7 points] Prove by induction that for all n N, 1/ S n. You will probably want to strengthen the induction hypothesis. You may not assume your generalization from the previous part is true; you must prove that statement if you wish to use it. Solution. Strengthen the theorem to P (n) = For all a/b S n with a, b N where a/b is in reduced form, b = 2 k for some k Clearly, if n P (n), then n 1/ S n. Thus, we need only prove n P (n). Base cases. 0 = 0/2 0, 1 = 1/2 0. Induction. Assume P (n). By the construction of S n+1, all elements of S n+1 are of the form (x + y)/2 for some x, y S n. Fix any two such elements x, y S n, where (by the induction hypothesis) x = a/2 k and y = c/2 l with a, b, k, l N. Assume WLOG k l. Then, (x + y)/2 = (a/2 k + c/2 l )/2 = (a/2 k + c2 k l /2 k )/2 = (a + c2 k l )/2 k+1 While this may not be in reduced form, reduction cannot introduce any non-2 factors into the denominator, so this fits the form specified by the theorem. Therefore, P (n + 1). By induction, n P (n). Thus, n 1/ S n. Problem 4 [10 points] Relations (a) [1 point] What is the common name for the reflexive closure of the less than relation? Solution. less than or equal to (b) [1 point] What is the common name for the transitive closure of the child of relation? Solution. descendant of (c) [2 points] What is the common name for the inverse of the teaches relation (that is, if xt y means x teaches y, how do we say at 1 b)? Solution. is taught by

4 Handout 26: Quiz 1 Solutions 4 Name: (d) [ points] Consider the partial order arb iff a divides b without remainder on the universe of natural numbers greater than 1. What are the minimal elements under this relation usually called? Solution. primes (e) [ points] Is the relation xry iff x = y or x is not y s roommate an equivalence relation? Why or why not? Solution. No, because it is not transitive. If x and y are roommates but neither is a roommate of z, then xrz and zry but xry. Problem 5 [10 points] Precedence Relations Death is planning the end of the world. This involves a number of tasks each of which takes one minute to complete. The prerequisites associated with these tasks are listed below. Abbrv. Task Prerequisites C Compose a requiem N Notify the UN B D Signal the daemons B B Blow the trumpet T Sell T-shirts: All I got was this lousy t-shirt. N Q Grade the quiz S G Open the gates D,N S Put out the sun C,N (a) [5 points] Represent the tasks and their prerequisites as a directed graph. Solution. This is very similar to the constructions in the notes. (b) [5 points] Death is omnipotent and can therefore work on as many tasks at a time as he wishes. What is the minimum amount of time required for him to end the world? Why? (An explanation without proof is fine.) Solution. 4 time units, since that is the length of the critical path. Problem 6 [5 points] Graphs Let G = (V, E) be a simple, undirected graph, and let C be a set of colors. Define a partial coloring of G using C to be a function that assigns to each node in V either a color in C or no color. (That is, it colors some, none, or all of the nodes using colors in C.) Define a partly colored graph to be a simple, undirected graph G = (V, E) together with a partial coloring of G. Prove the following fact about partly colored graphs:

5 Handout 26: Quiz 1 Solutions 5 Name: In a partly colored graph, any walk connecting a colored node to an uncolored node must include a colored node adjacent to an uncolored node. Solution. Consider the path p = v 1 v 2 v k, of which v 1 is colored and v k is uncolored. Assume there is no edge (v i, v i+1 ) such that v i is colored and v i+1 is uncolored. Then, by induction starting from v 1, v i are colored for all i, so v k is colored; this is a contradiction. Problem 7 [15 points] Algorithms Let G = (V, E) be a simple, undirected graph. The following algorithm, RedBlue, manipulates partial colorings of G using C = {red, blue}. (Partial colorings are defined in the previous problem.) Initially, one vertex r of G is colored red, and all the other vertices are uncolored. At each step, one of two events occurs: a) Some uncolored vertex v that is adjacent to a red vertex becomes blue. b) Some uncolored vertex v that is adjacent to a blue vertex becomes red. If neither rule can be applied, the algorithm terminates. (a) [ points] Formalize this algorithm as a state machine; that is, define Q, Q 0, and δ. Solution. A state q consists of a (total) function f from V to {red, blue, uncolored}. A start state is any state in which f maps exactly one v V to red, and all others to uncolored. The transitions are all of the form: (q, q ) δ v V q.f(v) = uncolored [ (q.f(v) = red w f(w) = blue (v, w) E) (q.f(v) = blue w q.f(w) = red (v, w) E) ] w v q.f(w) = q.f(w) (b) [ points] Prove that the algorithm eventually terminates. Solution. We describe a termination function. Define the value of the termination function to be the number d of uncolored vertices. Each step decreases d, so we can appeal to the Termination Theorem and conclude that the algorithm always terminates. Alternatively, we can expand the proof a bit more: since d starts at n 1 and is always 0, it must eventually reach some minimum value (by well-ordering), whereupon the algorithm terminates. (c) [ points] Using the fact stated in Problem 6, prove that if G is connected, then all vertices are colored when the algorithm terminates.

6 Handout 26: Quiz 1 Solutions 6 Name: Solution. Say the algorithm terminates with some vertex uncolored. Since the graph is connected, there is a path from v to some colored vertex w. By the given fact, there must be an edge along this path for which one endpoint x is colored and the other y is uncolored. We can apply one of the two steps above, assigning to y the opposite of x s color. Thus, the algorithm should not have terminated yet. (d) [ points] Consider applying this algorithm to bipartite graphs, i.e., graphs in which the vertices can be partitioned into two sets A and B such that there is no edge between any pair of vertices in A, and similarly for B. Suppose the node r that is colored red in the initial state is in set A. Using the Invariant Theorem, prove that in any reachable state, only vertices of A are colored red and only vertices of B are colored blue. Solution. The Invariant Theorem requires that we show some condition is true in all the start states, and that it is preserved by every step. P (n) = After n vertices are colored, all red vertices are in A and all blue vertices are in B. Base case. In any start state, there is exactly one red vertex in R by construction and no blue vertices. Therefore, P (1). Induction. Assume P (n). Then, consider an execution in which n + 1 vertices are colored. Consider the vertex i which was colored last. If we uncolor it, we can apply the induction hypothesis and conclude that all red vertices were in R and all blue vertices were in B. Now we must prove that vertex i is colored correctly. Assume WLOG that vertex i is in B. For i to have been colored, it must have been the case that i was adjacent to a colored vertex; but all vertices adjacent to i are in R (by the definition of bipartite), so i must have been colored blue. This satisfies the invariant theorem, so P (n + 1). Thus, n P (n), so all red vertices are in R and all blue vertices are in B. (e) [ points] Combine the previous parts to prove that when run on a connected bipartite graph, the above algorithm terminates and outputs a valid 2-coloring. Solution. We proved the algorithm terminates on connected graphs with all vertices colored. By the invariant above, this means all vertices in R are red and all vertices in B are blue. By the definition of bipartite, there exists no pair of like-colored vertices connected by an edge, so this is a valid 2-coloring.